Thank you. I made a computational mistake and now agree with your answer.
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joroMar 16 '12 at 9:15

Per juan's answer your sum is unconditionally $=\sum_{\rho} \rho^{-1}+\overline{\rho}^{-1}$ (the constant depends if one takes only Im(s)>0 or not). Since on the critical line the terms are equal does this mean the difference of the sums over potential Siegel zeros must vanish unconditionally?
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joroMar 16 '12 at 15:52

The series $\sum_\rho \rho^{-1}$ over the non-trivial zeros is not absolutely convergent, this is proved in Davenport p. 80. But as Davenport says and proves in
page 81-82 the series converges conditionally provided one groups together
the terms from $\rho$ and its conjugate $\overline{\rho}$. And the value of
the sum can be given, independently of RH, as the constant $-B$ where
$B = -\frac12 \gamma-1+\frac12\log4\pi$. (This value was known to Riemann, as Siegel
says in his paper about the Riemann Nachlass).

In Lagarias formula the sum is over the non trivial zeros repeated according to its multiplicity. In fact the formula is the Mittag-Leffler expansion of the meromorphic function $\frac{\hat\zeta'(s)}{\hat\zeta(s)}$ that has poles just at the non-trivial zeros and at $0$ and $1$
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juanMar 16 '12 at 8:44

Thank you. I made a computational mistake and now agree with Micah's answer.
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joroMar 16 '12 at 9:14

Looks like Micah's sum $\sum_{\rho} \rho^{-1}+(1-\rho)^{-1}$ is exactly $\sum_{\rho} \rho^{-1}+\overline{\rho}^{-1}$ (Davenport's $B$ sums only Im(s)>0 and this explains the factor of $2$. Since on the critical line the terms are equal does this mean the difference of the sums over potential Siegel zeros must vanish unconditionally?
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joroMar 16 '12 at 12:50

In fact the two sums are equal $\sum_\rho(\rho^{-1}+(1-\rho)^{-1})= \sum_\rho (\rho^{-1}+\overline{\rho}^{-1})$. But I (or Davenport) am speaking about $\zeta(s)$. The case of $L(\chi,s)$ is a little more complicate. In this case the zeros are symmetric with respect to the line $\sigma=\frac12$ but not respect the real axis. The determination of the constant $B(\chi)$ is recent, due to Vorhauer in 2006. Davenport do not include it. You must see the book by Montgomery and Vaughan, Chapter 10. In the sum you must consider the zeros of $L(s,\chi)$ and those of $L(s,\overline{\chi})$.
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juanMar 16 '12 at 17:54

Edit: I endorse Juan's answer to the original question. The sum $\displaystyle{\sum_{\rho} \tfrac{1}{|\rho|}}$, running over the non-trivial zeros $\rho$ of $\zeta(s)$, is known to diverge, so at best $\displaystyle{\sum_{\rho} \tfrac{1}{\rho}}$ is conditionally convergent so you cannot re-arrange the terms.

In your second to last displayed equation, you removed the assumption that the sum runs over pairs of zeros $\rho$ and $1-\rho$. So it seems that Lagarias' result can be used to evaluate the sum
$$ \sum_{\rho} \frac{1}{\rho (1{-}\rho)}.$$

As you observed, assuming the Riemann Hypothesis $1-\rho =\overline{\rho}$ for any non-trivial zero $\rho$ of $\zeta(s)$. This implies that

These identities are not mysterious. They are simply the fact that the Riemann Zeta function has a Weierstrass product like any other meromorphic function of finite exponential order. Note here that $f'/f$ is called logarithmic derivative for a reason;)

Then it follows immediately

1) Yes, the zeros of the completed Riemann zeta function are exactly the nontrivial ones.